Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{1}{4} n^{2}(n+1)^{2} $$

Short answer

The statement holds for all natural numbers n by mathematical induction.

Step-by-step solution

Base Case

Verify the statement for the initial value, which is usually 1. For n = 1,\[1^3 = \frac{1}{4} \cdot 1^2 \cdot (1+1)^2\]Simplify the right side:\[1^3 = \frac{1}{4} \cdot 1 \cdot 4 = 1\]Both sides are equal, thus the base case holds.

Inductive Hypothesis

Assume the statement is true for some natural number k, i.e.,\[1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{1}{4} k^2 (k+1)^2\]This is the inductive hypothesis.

Inductive Step

Show that the statement holds for k + 1. Consider\[1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3\]Using the inductive hypothesis,\[ \frac{1}{4} k^2 (k+1)^2 + (k+1)^3\]

Simplify the Expression

Factor out (k+1)^2 from the expression:\[ \frac{1}{4} k^2 (k+1)^2 + (k+1)^3 = (k+1)^2 \left( \frac{1}{4} k^2 + k+1 \right)\]

Further Simplification

Combine the terms inside the parenthesis:\[ (k+1)^2 \left( \frac{1}{4} k^2 + k + 1 \right) = (k+1)^2 \left( \frac{1}{4} k^2 + \frac{4k}{4} + \frac{4}{4} \right) = (k+1)^2 \left( \frac{1}{4} k^2 + \frac{4k}{4} + \frac{4}{4} \right) = (k+1)^2 \left( \frac{1}{4} (k^2 + 4k + 4) \right)\]Recognize that the term inside the parenthesis is a perfect square:\[= (k+1)^2 \cdot \frac{1}{4} (k+2)^2\]

Combine and Conclude

Therefore,\[ 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 = \frac{1}{4} (k+1)^2 (k+2)^2 \]This matches the form of the original statement with k+1 substituted for n, thus the statement holds for k+1.

Final Conclusion

Since both the base case and the inductive step have been proven, by the principle of mathematical induction, the statement is true for all natural numbers n.

Key concepts

algebraic proof

Algebraic proofs involve verifying mathematical statements using algebra. This type of proof uses equations and algebraic manipulations to demonstrate that statements are true. In this case, we are proving a formula for the sum of cubes of natural numbers using algebraic steps. We start by verifying that the formula holds for the initial value (base case). Then, we assume it's true for an arbitrary value (inductive hypothesis) before proving it for the next value (inductive step). This structured approach ensures each step logically flows from the previous one.

natural numbers

Natural numbers are the set of positive integers beginning from 1 and increasing without end: 1, 2, 3, 4, and so on. They are foundational in mathematical proofs, especially with the principle of mathematical induction. By proving a statement for the first natural number and then for n+1 assuming it is true for any natural number n, we show it's true for all natural numbers. It's essential to use natural numbers for induction because they have a well-defined start point and increase incrementally.

inductive hypothesis

The inductive hypothesis is a critical step in the principle of mathematical induction. When we assume that a statement is true for some arbitrary natural number k, it allows us to build upon this assumption. This hypothesis acts as a temporary truth which we use to demonstrate that if the statement holds for k, it will also hold for k+1. For example, we assume that \(1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{1}{4} k^2 (k+1)^2\) is true. This assumption lets us focus on proving the statement for k+1.

mathematical statement verification

Verifying a mathematical statement involves checking its validity through various methods, such as algebraic proof and induction. Initially, we verify the base case to ensure the statement works for the lowest natural number, typically 1. Then, by assuming it is true for an arbitrary number (inductive hypothesis), we move to show it’s true for the next number (k+1). For instance, verifying the formula \(1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^2 (n+1)^2\) involves confirming it holds for n=1, using the induction hypothesis for k, and proving it for k+1.